Problem: Is ${402179}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {402179}= &&{4}\cdot100000+ \\&&{0}\cdot10000+ \\&&{2}\cdot1000+ \\&&{1}\cdot100+ \\&&{7}\cdot10+ \\&&{9}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {402179}= &&{4}(99999+1)+ \\&&{0}(9999+1)+ \\&&{2}(999+1)+ \\&&{1}(99+1)+ \\&&{7}(9+1)+ \\&&{9} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {402179}= &&\gray{4\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {4}+{0}+{2}+{1}+{7}+{9} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${402179}$ is divisible by $3$ if ${ 4}+{0}+{2}+{1}+{7}+{9}$ is divisible by $3$ Add the digits of ${402179}$ $ {4}+{0}+{2}+{1}+{7}+{9} = {23} $ If ${23}$ is divisible by $3$ , then ${402179}$ must also be divisible by $3$ ${23}$ is not divisible by $3$, therefore ${402179}$ must not be divisible by $3$.